What data structures are used to solve Longest Common Subsequence?

This problem uses String, Dynamic Programming. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Longest Common Subsequence?

Insight 1: Recognizing that the problem can be solved using dynamic programming by building a table of solutions to subproblems. Insight 2: Understanding that the length of the LCS at `text1[i]` and `text2[j]` depends on whether `text1[i]` and `text2[j]` are equal or not. If they are equal, the LCS length is 1 plus the LCS length of `text1[0...i-1]` and `text2[0...j-1]`. If they are not equal, the LCS length is the maximum of the LCS length of `text1[0...i-1]` and `text2[0...j]` and the LCS length of `text1[0...i]` and `text2[0...j-1]`.

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Longest Common Subsequence - LeetCode 1143 Solution

Q: What is the time complexity of Longest Common Subsequence?

The optimal solution has a time complexity of O(m*n), where m and n are the lengths of text1 and text2, respectively. The code iterates through the entire dp array of size (m+1) x (n+1) once..

Q: What is the space complexity of Longest Common Subsequence?

The space complexity is O(m*n), where m and n are the lengths of text1 and text2, respectively. The code uses a 2D array `dp` of size (m+1) x (n+1) to store the intermediate results..

Q: Which companies ask LeetCode 1143. Longest Common Subsequence in interviews?

This problem is frequently asked at Accolite, Arista Networks, BP, ByteDance, Citadel and 4 more companies.

Longest Common Subsequence - Complete Solution Guide

Longest Common Subsequence is LeetCode problem 1143, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3.

Problem Statement

Given two strings text1 and text2 , return the length of their longest common subsequence . If there is no common subsequence , return 0 . A subsequence of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters. For example, "ace" is a subsequence of "abcde" . A common subsequence of two strings is a subsequence that is common to both strings. Example 1: Input: text1 = "abcde", text2 =

Detailed Explanation

The problem asks us to find the length of the longest common subsequence between two given strings, `text1` and `text2`. A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements. A common subsequence is a subsequence that is common to both input strings. The input consists of two strings, `text1` and `text2`, composed of lowercase English characters. The output is an integer representing the length of their longest common subsequence. If there is no common subsequence, we return 0. The length of each input string is between 1 and 1000.

Solution Approach

The provided code implements a dynamic programming approach to solve the Longest Common Subsequence problem. It constructs a 2D array `dp` where `dp[i][j]` stores the length of the LCS of `text1[0...i-1]` and `text2[0...j-1]`. The array is built iteratively, starting from the base case of empty prefixes (`dp[0][j]` and `dp[i][0]` are initialized to 0). If `text1[i-1]` and `text2[j-1]` are equal, it means that we can extend the LCS by one character, so `dp[i][j]` is set to `dp[i-1][j-1] + 1`. If they are not equal, we take the maximum of the LCS lengths obtained by either excluding the last character of `text1` or excluding the last character of `text2`. Finally, the length of the LCS of the entire strings `text1` and `text2` is stored in `dp[m][n]`, where `m` and `n` are the lengths of `text1` and `text2`, respectively.

Step-by-Step Algorithm

Step 1: Initialize a 2D array `dp` of size `(m+1) x (n+1)` with all elements set to 0, where `m` is the length of `text1` and `n` is the length of `text2`. The extra row and column are used to handle the base case of empty prefixes.
Step 2: Iterate through the `dp` array starting from index `(1, 1)` to `(m, n)`. For each cell `dp[i][j]`, compare the characters `text1[i-1]` and `text2[j-1]`.
Step 3: If `text1[i-1]` is equal to `text2[j-1]`, it means that the characters contribute to the LCS. Set `dp[i][j]` to `dp[i-1][j-1] + 1`.
Step 4: If `text1[i-1]` is not equal to `text2[j-1]`, it means that we need to consider the LCS without including either `text1[i-1]` or `text2[j-1]`. Set `dp[i][j]` to `max(dp[i-1][j], dp[i][j-1])`.
Step 5: After iterating through the entire `dp` array, the value `dp[m][n]` will contain the length of the LCS of `text1` and `text2`. Return this value.

Key Insights

Insight 1: Recognizing that the problem can be solved using dynamic programming by building a table of solutions to subproblems.
Insight 2: Understanding that the length of the LCS at `text1[i]` and `text2[j]` depends on whether `text1[i]` and `text2[j]` are equal or not. If they are equal, the LCS length is 1 plus the LCS length of `text1[0...i-1]` and `text2[0...j-1]`. If they are not equal, the LCS length is the maximum of the LCS length of `text1[0...i-1]` and `text2[0...j]` and the LCS length of `text1[0...i]` and `text2[0...j-1]`.
Insight 3: Using a 2D array to store the lengths of LCS for all possible prefixes of the two strings avoids redundant calculations.

Complexity Analysis

Time Complexity: O(m*n), where m and n are the lengths of text1 and text2, respectively. The code iterates through the entire dp array of size (m+1) x (n+1) once.

Space Complexity: O(m*n), where m and n are the lengths of text1 and text2, respectively. The code uses a 2D array `dp` of size (m+1) x (n+1) to store the intermediate results.

Topics

This problem involves: String, Dynamic Programming.

Companies

Asked at: Accolite, Arista Networks, BP, ByteDance, Citadel, DoorDash, Nutanix, Optum, tcs.

Longest Common Subsequence | LeetCode 1143

Medium

String Dynamic Programming

Time: O(m*n), where m and n are the lengths of text1 and text2, respectively. The code iterates through the entire dp array of size (m+1) x (n+1) once.Space: O(m*n), where m and n are the lengths of text1 and text2, respectively. The code uses a 2D array `dp` of size (m+1) x (n+1) to store the intermediate results.

Solution Code

Available in:

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Problem Info

Description

Problem 1143: Longest Common Subsequence - Detailed explanation and solution approaches.

Topics

String Dynamic Programming

Companies

AccoliteArista NetworksBPByteDanceCitadelDoorDashNutanixOptum

Stats

Difficulty:Medium

Languages:2

Companies:9

Problem #:1143

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Initialize a 2D array `dp` of size `(m+1) x (n+1)` with all elements set to 0, where `m` is the length of `text1` and `n` is the length of `text2`. The extra row and column are used to handle the base case of empty prefixes.
2
Step 2: Iterate through the `dp` array starting from index `(1, 1)` to `(m, n)`. For each cell `dp[i][j]`, compare the characters `text1[i-1]` and `text2[j-1]`.
3
Step 3: If `text1[i-1]` is equal to `text2[j-1]`, it means that the characters contribute to the LCS. Set `dp[i][j]` to `dp[i-1][j-1] + 1`.
4
Step 4: If `text1[i-1]` is not equal to `text2[j-1]`, it means that we need to consider the LCS without including either `text1[i-1]` or `text2[j-1]`. Set `dp[i][j]` to `max(dp[i-1][j], dp[i][j-1])`.
5
Step 5: After iterating through the entire `dp` array, the value `dp[m][n]` will contain the length of the LCS of `text1` and `text2`. Return this value.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

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Complexity Analysis

⏱️

Time Complexity

O(m*n), where m and n are the lengths of text1 and text2, respectively. The code iterates through the entire dp array of size (m+1) x (n+1) once.

💾

Space Complexity

O(m*n), where m and n are the lengths of text1 and text2, respectively. The code uses a 2D array `dp` of size (m+1) x (n+1) to store the intermediate results.

📝

Why?

The time complexity is O(m*n) because we have nested loops iterating over each cell in the `dp` matrix. The space complexity is O(m*n) because we store the LCS lengths of all prefix combinations in the `dp` matrix.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursive Approach with Memoization

A recursive solution can be implemented with memoization to avoid redundant calculations. This approach explores the possible subsequences recursively and stores the results in a memo table.

Time: O(m*n), where m and n are the lengths of text1 and text2.Space: O(m*n) due to the memoization table and the recursive call stack.

Trade-offs: While having the same time complexity, the recursive approach might be slightly less efficient due to function call overhead. However, it can be more intuitive for some programmers to understand the recursive logic. The space complexity is equivalent, as memoization essentially creates the same table as the DP approach.

Space Optimized Dynamic Programming

The space complexity of the standard dynamic programming approach can be reduced to O(min(m, n)) by using only two rows of the `dp` table at a time. This optimization works because the current row's value only depends on the previous row.

Time: O(m*n), where m and n are the lengths of text1 and text2.Space: O(min(m, n)).

Trade-offs: This approach optimizes space but makes the code potentially harder to read and debug. It's beneficial when memory is a major constraint.

✨

Key Insights

1Insight 1: Recognizing that the problem can be solved using dynamic programming by building a table of solutions to subproblems.
2Insight 2: Understanding that the length of the LCS at `text1[i]` and `text2[j]` depends on whether `text1[i]` and `text2[j]` are equal or not. If they are equal, the LCS length is 1 plus the LCS length of `text1[0...i-1]` and `text2[0...j-1]`. If they are not equal, the LCS length is the maximum of the LCS length of `text1[0...i-1]` and `text2[0...j]` and the LCS length of `text1[0...i]` and `text2[0...j-1]`.
3Insight 3: Using a 2D array to store the lengths of LCS for all possible prefixes of the two strings avoids redundant calculations.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: One or both of the input strings are empty. The code handles this gracefully because the `dp` array is initialized with 0s. If either string is empty, the LCS will be 0, which is the correct answer.
Edge case 2: The two strings are identical. The algorithm correctly identifies the LCS as the entire string and returns its length.
Edge case 3: The two strings have no common characters. The algorithm correctly identifies that the LCS is 0 and returns this value.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not properly initializing the `dp` array, leading to incorrect base cases.

✗Common mistake 2: Incorrectly implementing the recurrence relation, especially when the characters don't match.

✗Common mistake 3: Not understanding the difference between a subsequence and a substring. This problem requires finding a subsequence, where characters don't have to be contiguous.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Bioinformatics - Finding similarities between DNA sequences. The LCS can be used to identify conserved regions in different DNA sequences, which may indicate important functional elements.

🏦Real-world scenario 2: Text editing - The 'diff' utility, which highlights differences between two versions of a text file, utilizes the concept of LCS. The LCS identifies the common parts of the two files, and the differences are then highlighted based on what is not part of the LCS.

🎮Real-world scenario 3: Speech recognition - Determining the similarity between the spoken words and the words in a dictionary. The LCS can be used to find the best match for the spoken words in the dictionary.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 3, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

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Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Longest Common Subsequence"?

For this medium-level String problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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